Question: Simplify and expand the following expression: $ \dfrac{3q - 6}{q - 3}+\dfrac{-4}{q + 9} $
Solution: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(q - 3)(q + 9)$ Multiply the first term by $\dfrac{q + 9}{q + 9}$ $ \begin{align*} \dfrac{3q - 6}{q - 3} \times \dfrac{q + 9}{q + 9} & = \dfrac{(3q - 6)(q + 9)}{(q - 3)(q + 9)} \\ & = \dfrac{3q^2 + 21q - 54}{(q - 3)(q + 9)}\end{align*} $ Multiply the second term by $\dfrac{q - 3}{q - 3}$ $ \begin{align*} \dfrac{-4}{q + 9} \times \dfrac{q - 3}{q - 3} & = \dfrac{(-4)(q - 3)}{(q + 9)(q - 3)} \\ & = \dfrac{-4q + 12}{(q + 9)(q - 3)}\end{align*} $ Now we have: $ = \dfrac{3q^2 + 21q - 54}{(q - 3)(q + 9)} + \dfrac{-4q + 12}{(q + 9)(q - 3)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{3q^2 + 21q - 54 - 4q + 12}{(q - 3)(q + 9)} $ $ = \dfrac{3q^2 + 17q - 42}{(q - 3)(q + 9)}$ Expand the denominator: $ = \dfrac{3q^2 + 17q - 42}{q^2 + 6q - 27}$